Operator Taylor series
mike3
I saw this:
http://en.wikipedia.org/wiki/Finite_difference#Calculus_of_finite_differences
which shows that the "forward difference operator" can be expressed as
a "Taylor series of operators" in the differentiation operator:
Delta_h = hD + hD^2/2! + hD^3/3! + ... = "exp(hD) - 1"
How does one come up with this (it mentions "applying Taylor's theorem
to h", but I'm not sure how exactly that'd be done here for
operators)? Also, is there a similar expression for the summation
operator Delta^-1_h in terms of powers of the integration operator? If
so, what is it?
mike3
>
> which shows that the "forward difference operator" can be expressed as
> a "Taylor series of operators" in the differentiation operator:
>
> Delta_h = hD + hD^2/2! + hD^3/3! + ... = "exp(hD) - 1"
>
> How does one come up with this (it mentions "applying Taylor's theorem
> to h", but I'm not sure how exactly that'd be done here for
> operators)? Also, is there a similar expression for the summation
> operator Delta^-1_h in terms of powers of the integration operator? If
> so, what is it?
I noticed there's a formula for the sum as another series of
differentiation
operators (for h = 1):
Delta^-1 = I - 1/2 + 1/12 D - 1/720 D^3 + 1/30240 D^5 - 1/129600 D^7
+ 1/47900160 D^9 - 691/1307674368000 D^11 + ...
where I is the integration operator. However this does not converge
for
all functions (take an analytic function in there whose derivatives at
a point
increase factorially.). Is there some analogue of "analytic
continuation of an
operator" that could be employed here? :) Is there also an integral
series
as I mentioned?
Robert Israel
> Hi.
>
> I saw this:
>
>
>http://en.wikipedia.org/wiki/Finite_difference#Calculus_of_finite_differences
>
>
> which shows that the "forward difference operator" can be expressed as
> a "Taylor series of operators" in the differentiation operator:
>
> Delta_h = hD + hD^2/2! + hD^3/3! + ... = "exp(hD) - 1"
>
> How does one come up with this (it mentions "applying Taylor's theorem
> to h", but I'm not sure how exactly that'd be done here for
> operators)?
Taylor's theorem says (if f is analytic and h sufficiently small)
f(x + h) = sum_{j=0}^infty (D^j f)(x) h^j/j!
so
(Delta_h f)(x) = f(x + h) - f(x) = sum_{j=1}^infty (D^j f)(x) h^j/j!
= (sum_{j=1}^infty (h^j/j!) D^j)(f)(x)
= (exp(hD) - 1)(f)(x)
> Also, is there a similar expression for the summation
> operator Delta^-1_h in terms of powers of the integration operator? If
> so, what is it?
Delta_h doesn't really have an inverse, and neither does D. But proceeding
formally,
(Delta_h)^(-1) = h^(-1) D^(-1) - 1/2 + h D/12 - ...
= h^(-1) D^(-1) + sum_{n=1}^infty (B_n h^(n-1)/n!) D^(n-1)
where B_n is the n'th Bernoulli number. This is the Euler-Maclaurin summation
formula.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
mike3
<isr...@math.MyUniversitysInitials.ca> wrote:
> mike3 <mike4...@yahoo.com> writes:
<snip>
> Taylor's theorem says (if f is analytic and h sufficiently small)
>
> f(x + h) = sum_{j=0}^infty (D^j f)(x) h^j/j!
>
> so
>
> (Delta_h f)(x) = f(x + h) - f(x) = sum_{j=1}^infty (D^j f)(x) h^j/j!
> = (sum_{j=1}^infty (h^j/j!) D^j)(f)(x)
> = (exp(hD) - 1)(f)(x)
>
Ahh. That makes sense.
> > Also, is there a similar expression for the summation
> > operator Delta^-1_h in terms of powers of the integration operator? If
> > so, what is it?
>
> Delta_h doesn't really have an inverse, and neither does D. But proceeding
> formally,
>
> (Delta_h)^(-1) = h^(-1) D^(-1) - 1/2 + h D/12 - ...
> = h^(-1) D^(-1) + sum_{n=1}^infty (B_n h^(n-1)/n!) D^(n-1)
> where B_n is the n'th Bernoulli number. This is the Euler-Maclaurin summation
> formula.
This seems to be the formula I posted in the 2nd post.
So what about that question of "analytic continuation of operators"?
E.g. to "extend" the operator defined by this "Taylor series" to a
wider
domain of functions (it doesn't work on, e.g. functions that are worse
than
exponential type -- try it on e^e^x and see what happens)? Note that a
function
defined a by Taylor series with limited area of convergence may be
extended
to a larger range of input values via analytic continuation. Is
something
similar possible with an operator defined as a "Taylor series" in
terms of
some other operator which has a wider domain (e.g. you can apply D^n
to
e^e^x)?
And also, what about the other, i.e. could some summation operator be
given
as powers of "some" integration operator?
mike3
> On Dec 21, 5:18 pm, Robert Israel
>
> <isr...@math.MyUniversitysInitials.ca> wrote:
> > mike3 <mike4...@yahoo.com> writes:
> <snip>
>
> > f(x + h) = sum_{j=0}^infty (D^j f)(x) h^j/j!
>
> > so
>
> > (Delta_h f)(x) = f(x + h) - f(x) = sum_{j=1}^infty (D^j f)(x) h^j/j!
> > = (sum_{j=1}^infty (h^j/j!) D^j)(f)(x)
> > = (exp(hD) - 1)(f)(x)
>
> Ahh. That makes sense.
>
> > > Also, is there a similar expression for the summation
> > > operator Delta^-1_h in terms of powers of the integration operator? If
> > > so, what is it?
>
> > Delta_h doesn't really have an inverse, and neither does D. But proceeding
> > formally,
>
> > (Delta_h)^(-1) = h^(-1) D^(-1) - 1/2 + h D/12 - ...
> > = h^(-1) D^(-1) + sum_{n=1}^infty (B_n h^(n-1)/n!) D^(n-1)
> > where B_n is the n'th Bernoulli number. This is the Euler-Maclaurin summation
> > formula.
>
> This seems to be the formula I posted in the 2nd post.
>
> E.g. to "extend" the operator defined by this "Taylor series" to a
> wider
> domain of functions (it doesn't work on, e.g. functions that are worse
> than
> exponential type -- try it on e^e^x and see what happens)? Note that a
> function
> defined a by Taylor series with limited area of convergence may be
> extended
> something
> similar possible with an operator defined as a "Taylor series" in
> terms of
> some other operator which has a wider domain (e.g. you can apply D^n
> to
> e^e^x)?
>
> And also, what about the other, i.e. could some summation operator be
> given
> as powers of "some" integration operator?
Any answers?
Gottfried Helms
a bit late - but since I'm not familiar with the
common use of the "operator"-term in function-analysis
I step aside if I see it mentioned, especially, if it
is involved in operations like you mention.
But anyway - I tried to translate this to my operator-concept
of matrix-operators on formal powerseries. There we have
the (upper triabular) Pascal-matrix, which performs on a vandermonde
vector V(x) = rowvector(1,x,x^2,x^3,x^4,x...)
V(x) * P = V(x+1)
which apparently mimics that Delta_h (h=1) of the functions-analysis.
(note, that I used the transposed V and P here compared to my usual notation
in tetration-forum etc for notational ease)
Moreover with a matrix-operator F implementing a function f(x) on
a vandermondevector
V(x) * F = V(f(x))
we have also
V(x) * P * F = V(f(x+1))
and can read this as
PF = P * F
as application of the "differenceoperator" at a "functionoperator"
giving a new object.
-------------------------
Now the matrix(operator) P can be written as a matrix-exponential
P = exp(L) = I + L + L^2/2! + L^3/3! + ...
where L is the first upper subdiagonal containing
updiag(1)(1,2,3,4,..)
Also we have
V(x)* L* F = row(0,f(x)', (f(x)^2)', (f(x)^3)', ...
= (V(x))' \\ f' meaning the derivative
So again this seems to match the hD-notation above.
The inverse operation, the (formal) integral on powerseries,
can be expressed by the transposed of L, doubly factorially
rescaled:
M = f^2 * L~ * F^2
where F=diag(0!,1!,2!,3!,...) and f = F^-1 and
M = lowdiag(1)(1,1/2,1/3,1/4,...)
Then
V(x) * M F = V(integral(f(x)))
However, M cannot be fully understood as "the inverse" of L because
M*L = diag(0, 1,1,1,1,..) \\ having leading zero
and
L*M has a trailing zero if the matrices are finite.
Gottfried
==========================================================================
Codeexample Pari/GP, using functions from PariTTY
dim=6
\\ ---------------
V(x) = vector(dim,r,x^(r-1)) \\ rowvector
F = matdiagonal(vector(dim,r,(r-1)!))
f = F^-1
P = matpascal(dim-1)~
fS2F = f*makemat_stirling2(dim)*F \\ implements operator for exp(x)-1
MEpx(M) = \\ implementation of matrixexponential of M
MLog(M) = \\ implementation of matrixlogarithm of M
\\ ----------------
V(x)
% [1, x, x^2, x^3, x^4, x^5]
V(x) * P
% [1,
x + 1,
x^2 + 2*x + 1,
x^3 + 3*x^2 + 3*x + 1,
x^4 + 4*x^3 + 6*x^2 + 4*x + 1,
x^5 + 5*x^4 + 10*x^3 + 10*x^2 + 5*x + 1 ]
\\ = [ 1, x+1, (x+1)^2 , (x+1)^3, ...]
\\ = V(x+1)
%pri L=MLog(P)
% L:
0 1 0 0 0 0
. 0 2 0 0 0
. . 0 3 0 0
. . . 0 4 0
. . . . 0 5
. . . . . 0
V(x) * fS2F
% [1,
1/120*x^5 + 1/24*x^4 + 1/6*x^3 + 1/2*x^2 + x,
1/4*x^5 + 7/12*x^4 + x^3 + x^2,
5/4*x^5 + 3/2*x^4 + x^3,
2*x^5 + x^4,
x^5]
\\ for lim dim->inf
\\ = row( 1 , exp(x)-1 , (exp(x)-1)^2, ,...)
V(x)*L*fS2F
% [0,
1/24*x^4 + 1/6*x^3 + 1/2*x^2 + x + 1,
5/4*x^4 + 7/3*x^3 + 3*x^2 + 2*x,
25/4*x^4 + 6*x^3 + 3*x^2,
10*x^4 + 4*x^3,
5*x^4]
\\ for lim dim->inf
\\ = row( 1 , (exp(x)-1)' , ((exp(x)-1)^2)', ,...)
\\ = V( exp(-1)) '
%pri M = f^2 * L~ * F^2
0 . . . . .
1 0 . . . .
0 1/2 0 . . .
0 0 1/3 0 . .
0 0 0 1/4 0 .
0 0 0 0 1/5 0
V(x)*M*fS2F
% [x,
1/120*x^5 + 1/24*x^4 + 1/6*x^3 + 1/2*x^2,
7/60*x^5 + 1/4*x^4 + 1/3*x^3,
3/10*x^5 + 1/4*x^4,
1/5*x^5,
0]
\\ for lim dim->inf
\\ = [ x, int(exp(x)-1)dx, int( (exp(x)-1)^2)dx, ... ]
\\ = int(V(exp(x)-1))dx
%pri M*L
0 . . . . .
. 1 . . . .
. . 1 . . .
. . . 1 . .
. . . . 1 .
. . . . . 1
%pri L*M
1 . . . . .
. 1 . . . .
. . 1 . . .
. . . 1 . .
. . . . 1 .
. . . . . 0
====================================================
Gottfried Helms
Am 02.01.2010 15:02 schrieb Gottfried Helms:
(...)
> But anyway - I tried to translate this to my operator-concept
> of matrix-operators on formal powerseries. There we have
> the (upper triabular) Pascal-matrix, which performs on a vandermonde
*********** triangular
> M = f^2 * L~ * F^2
>
> where F=diag(0!,1!,2!,3!,...) and f = F^-1 and
> M = lowdiag(1)(1,1/2,1/3,1/4,...)
>
> Then
> V(x) * M F = V(integral(f(x)))
**** V(x) * M F = integral V(f(x)) dx
Sorry -
Gottfried
Gottfried Helms
Hi Mike -
a bit late - but since I'm not familiar with the
common use of the "operator"-term in function-analysis
I step aside if I see it mentioned, especially, if it
is involved in operations like you mention.
But anyway - I tried to translate this to my operator-concept
of matrix-operators on formal powerseries. There we have
the (upper triangular) Pascal-matrix, which performs on a vandermonde
vector V(x) = rowvector(1,x,x^2,x^3,x^4,x^5,...)
V(x) * P = V(x+1)
which apparently mimics that Delta_h (h=1) of the functions-analysis.
(note, that I used the transposed V and P here compared to my usual notation
in tetration-forum etc for notational ease)
Moreover with a matrix-operator F implementing a function f(x) on
a vandermondevector
V(x) * F = V(f(x))
we have also
V(x) * P * F = V(f(x+1))
and can read this as
PF = P * F
as application of the "differenceoperator" at a "functionoperator"
giving a new object.
-------------------------
Now the matrix(operator) P can be written as a matrix-exponential
P = exp(L) = I + L + L^2/2! + L^3/3! + ...
where L is the first upper subdiagonal containing
updiag(1)(1,2,3,4,..)
Also we have
V(x)* L* F = row(0,f(x)', (f(x)^2)', (f(x)^3)', ...
\\ f()' meaning the derivative
\\ = diff V(x) / dx
So again this seems to match the hD-notation above.
The inverse operation, the (formal) integral on powerseries,
can be expressed by the transposed of L, doubly factorially
rescaled:
M = fac^2 * L~ * FAC^2
where FAC=diag(0!,1!,2!,3!,...) and fac = FAC^-1 and
M = lowdiag(1)(1,1/2,1/3,1/4,...)
Then
V(x) * M F = integral V(f(x)) dx
However, M cannot be fully understood as "the inverse" of L because
M*L = diag(0, 1,1,1,1,..) \\ having leading zero
and
L*M has a trailing zero if the matrices are finite.
Gottfried
==========================================================================
Codeexample Pari/GP, using functions from PariTTY
dim=6
\\ ---------------
V(x) = vector(dim,r,x^(r-1)) \\ rowvector
FAC = matdiagonal(vector(dim,r,(r-1)!))
fac = FAC^-1
P = matpascal(dim-1)~
fS2F = fac*makemat_stirling2(dim)*FAC \\ implements operator for exp(x)-1
%pri M = fac^2 * L~ * FAC^2
mike3
<snip>
So how can we get an inverse of the difference operator this
way? Can it take an inverse difference of log(1+x)? If so,
it should yield log(gamma(1+x)).
Rob Johnson
Indeed. I have more of the formal argument at
<http://www.whim.org/nebula/math/eulermac.html>.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Gottfried Helms
Hmm, first I need to know, whether and in which extent your
input and my interpretation in matrix-terms agree. Do they?
Then, if this should be the case, if the difference-operator
really can be seen/interpreted as pascalmatrix P - then the
inverse is nearly trivial.
So I think, there is some more here: because you state it as
difficult you may think in terms of
(P - I) * F
and using the vandermonde-notation
V(x) * (P-I) * F = V(f(x+1)) - V(f(x))
which performs in functional notation
x -> f(x+1) - f(x)
Is it this, what you really meant? Then, true, (P-I) cannot
be conventionally inverted.
But there is some workaround, and indeed then the bernoulli-
numbers come into the play. In my article on "sums of like
powers" [*1] I discussed such a workaround and came to the
invention of the ZETA-matrix, which implements the integrals
of the bernoulli-polynomials and performs summing of like
powers exploiting differences of the hurwitz-zeta-function
h_zeta(n,a) - h_zeta(n,b+1) = a^n+(a+1)^n + ... + b^n
ZETA is not triangular, and thus cannot be inverted directly.
However, it is intimately related to (P-I) and some normalizing
can make it triangular and invertible. Then we get a shifted
version of (P-I). (and possibly arrive in the near of the
required notion of a delta_h_inverse)
But all this is speculation unless and before we agree, that the
implied translation of your problem into this matrix-lingo is
meaningful at all, and that it also matches your question
in some way. (Then I'd really like to extend these considerations
further to make them fruitful)
Iterated cordially- ;-)
Gottfried
[*1] http://go.helms-net.de/math/binomial_new/04_3_SummingOfLikePowers.pdf
chap 3, pg 14
mike3
> Am 03.01.2010 11:01 schrieb mike3:
>
> > On Jan 2, 7:02 am, Gottfried Helms <he...@uni-kassel.de> wrote:
> > <snip>
>
> > So how can we get an inverse of the difference operator this
> > way? Can it take an inverse difference of log(1+x)? If so,
> > it should yield log(gamma(1+x)).
>
> Hmm, first I need to know, whether and in which extent your
> input and my interpretation in matrix-terms agree. Do they?
>
> Then, if this should be the case, if the difference-operator
> really can be seen/interpreted as pascalmatrix P - then the
> inverse is nearly trivial.
> So I think, there is some more here: because you state it as
> difficult you may think in terms of
> (P - I) * F
> and using the vandermonde-notation
>
> V(x) * (P-I) * F = V(f(x+1)) - V(f(x))
>
> which performs in functional notation
> x -> f(x+1) - f(x)
>
> Is it this, what you really meant? Then, true, (P-I) cannot
> be conventionally inverted.
Yeah, that's the operator I meant.
> But there is some workaround, and indeed then the bernoulli-
> numbers come into the play. In my article on "sums of like
> powers" [*1] I discussed such a workaround and came to the
> invention of the ZETA-matrix, which implements the integrals
> of the bernoulli-polynomials and performs summing of like
> powers exploiting differences of the hurwitz-zeta-function
>
> h_zeta(n,a) - h_zeta(n,b+1) = a^n+(a+1)^n + ... + b^n
>
> ZETA is not triangular, and thus cannot be inverted directly.
> However, it is intimately related to (P-I) and some normalizing
> can make it triangular and invertible. Then we get a shifted
> version of (P-I). (and possibly arrive in the near of the
> required notion of a delta_h_inverse)
>
> But all this is speculation unless and before we agree, that the
> implied translation of your problem into this matrix-lingo is
> meaningful at all, and that it also matches your question
> in some way. (Then I'd really like to extend these considerations
> further to make them fruitful)
>
> Iterated cordially- ;-)
>
What does that last part mean? How would one determine if it is
"meaningful at all"? It seems to make some sense, if that's what you
mean.
Gottfried Helms
> I noticed there's a formula for the sum as another series of
> differentiation
> operators (for h = 1):
>
> Delta^-1 = I - 1/2 + 1/12 D - 1/720 D^3 + 1/30240 D^5 - 1/129600 D^7
> + 1/47900160 D^9 - 691/1307674368000 D^11 + ...
>
> where I is the integration operator. However this does not converge
> for
Question: the coefficients mult by factorials give zeta-values
but at D^7 this differs. Is the value correct or is there some typo?
If the values are correct, please provide some more terms or the generation
rule.
Gottfried
Gottfried Helms
> differentiation
> operators (for h = 1):
>
> Delta^-1 = I - 1/2 + 1/12 D - 1/720 D^3 + 1/30240 D^5 - 1/129600 D^7
> + 1/47900160 D^9 - 691/1307674368000 D^11 + ...
>
> where I is the integration operator. However this does not converge
> for
> all functions (take an analytic function in there whose derivatives at
> a point
> increase factorially.). Is there some analogue of "analytic
> continuation of an
> operator" that could be employed here? :) Is there also an integral
> series
> as I mentioned?
I recognize this series now. It matches perfectly the application
of the ZETA-matrix in my matrix-concept.
---- recall definitions from other post --------
We use, as in the other post,
L as matrix-log of P (upper triangular pascalmatrix)
L = MLog(P) = updiag(1)(1,2,3,4,5,...)
and also as representation of the differential-operator
such that with a matrix-operator F
V(x) * F = V(f(x))
and
V(x) * L * F = d (V(f(x)) / dx
= V(f(x))'
and M the integrating operator
M = lowdiag(1)(1,1/2,1/3,...)
= fac^2 * L~ * FAC^2
such that
V(x) * M * F = int V(f(x)) dx
---------------------------------------------------------
I presented the ZETA-matrix as somehow representative for
the impossible inversion of the operator (P - I)^-1
With the above we can also describe the construction of
the ZETA-matrix as
Z = M
+ I*zeta(0) - zeta(-1)/1! * L + zeta(-2)/2! * L^2 - zeta(-3)/3! * L^3 + ... - ...
in Pari/GP
Z = M + sum(k=0,dim,(-1)^k*L^k*zetafrac(-k)/k!)
where Z is simply a differently normed version of my ZETA-matrix:
Z = - (ZETA*P^-1) (for the exact relation see below)
Now we describe the ZETA-matrix simply as sum of its upper subdiagonals,
which, by construction are weighted powers of L, just according to
the above formula for Z.
The additional first lower subdiagonal is M, and its occurence
is explained in my sum-of-like-powers article. There it was introduced
as effect of the infinite sum of reciprocals (=zeta(1)) divided by gamma(0),
which occurs in the infinite sum of powers of P, for which (P-I)^-1 is
a formal, non-evaluatable, expression.
So your mentioned series can be explained by the same logic with
which I explained the construction of the ZETA-matrix.
(See, for a comparision of Z and my ZETA, document at end of this post)
Taken as operator in the matrix-notation, we would use it like this:
V(1) * Z = [ 1/2 1/12 0 -1/120 0 1/252 0 -1/240 0 1/132 ,...)
= -rowvector( zeta(0), zeta(-1), zeta(-2), zeta(-3), ...)
= -rowvector( hzeta(0,1), hzeta(-1,1), hzeta(-2,1), hzeta(-3,1), ...)
V(2) * Z = [3/2, 13/12, 1, 119/120, 1, 253/252, 1, 239/240, ...]
= -rowvector( zeta(0)-1, zeta(-1)-1, zeta(-2)-1, zeta(-3)-1, ...)
= -rowvector( hzeta(0,2), hzeta(-1,2), hzeta(-2,2), hzeta(-3,2), ...)
V(x) * Z = -rowvector( hzeta(0,x), hzeta(-1,x), hzeta(-2,x), hzeta(-3,x), ...)
(V(3)-V(1)) * Z = [2, 3, 5, 9, 17, 33, 65, 129, ...]
= [1+2^0, 1+2, 1+2^2, 1+2^3, 1+2^4, ...]
(V(x+1)-V(x)) * Z = [1, x, x^2, x^3, x^4, x^5, x^6, x^7,...]
and with an additional matrixoperator F we had the according finite sums or
infinite series of f(x+k) with increasing k
V(x) * Z * F = resultvector
resultvector = - rowvector( ? ,
, f(x) + f(x+1) + f(x+2) + ...
, f(x)^2 + f(x+1)^2 + f(x+2)^2 + ...
, f(x)^3 + f(x+1)^3 + f(x+2)^3 + ...
... )
and should also have, (adapting the formula to get positive signs),
(V(x+1) -V(x))* Z * F = rowvector(1,f(x),f(x)^2, ...)
For instance, F being the matrixoperator for x -> exp(x) ,
namely F = fS2F*P \\ P being the upper triangular version!
(V(3) -V(2))* Z * F = [1.000000, 7.389056, 54.59815, 403.4288, 2980.958, ...]
= [ exp(2)^0, exp(2), exp(2)^2, ...]
= V(exp(2))
(V(3) -V(1))* Z * F = [2.000000, 10.10734, 61.98721, 423.5143, 3035.556, .,...]
= [ exp(1)^0 + exp(2)^0, exp(1)+ exp(2), exp(1)^2 + exp(2)^2, ...]
= V(exp(1)) + V(exp(2))
and this allows to implement the sum of functions with consecutive argument
f(m) + f(m+1) + f(m+2) + ... + f(n)
at least for integral interval n-m
-----------------------
What happens for fractional interval n-m? Examples:
(V(3) -V(2.5))* Z * F = [0.5000000, 4.599387, 39.91445, 329.8331, ...]
(V(2) -V(1.5))* Z * F = [0.5000000, 1.692020, 5.401833, 16.42142, ...]
(V(1) -V(0.5))* Z * F = [0.5000000, 0.6224593, 0.7310586, 0.8175745, ...]
Don't know, what this means, at the moment.
--------------------------
Does this all resemble your two questions so far correctly?
Gottfried
----------------------------------------------------------------------
Z = \\ according to the series in your post
-1/2 1/12 0 -1/120 0 1/252 0 -1/240
1 -1/2 1/6 0 -1/30 0 1/42 0
0 1/2 -1/2 1/4 0 -1/12 0 1/12
0 0 1/3 -1/2 1/3 0 -1/6 0
0 0 0 1/4 -1/2 5/12 0 -7/24
0 0 0 0 1/5 -1/2 1/2 0
0 0 0 0 0 1/6 -1/2 7/12
0 0 0 0 0 0 1/7 -1/2
ZETA~ = \\ my standard implementation in "summing-of-like-powers"
-1/2 -1/12 0 1/120 0 -1/252 0 1/240
-1 -1/2 -1/6 0 1/30 0 -1/42 0
0 -1/2 -1/2 -1/4 0 1/12 0 -1/12
0 0 -1/3 -1/2 -1/3 0 1/6 0
0 0 0 -1/4 -1/2 -5/12 0 7/24
0 0 0 0 -1/5 -1/2 -1/2 0
0 0 0 0 0 -1/6 -1/2 -7/12
0 0 0 0 0 0 -1/7 -1/2
Gottfried Helms
>
> What happens for fractional interval n-m? Examples:
>
> (V(3) -V(2.5))* Z * F = [0.5000000, 4.599387, 39.91445, 329.8331, ...]
> (V(2) -V(1.5))* Z * F = [0.5000000, 1.692020, 5.401833, 16.42142, ...]
> (V(1) -V(0.5))* Z * F = [0.5000000, 0.6224593, 0.7310586, 0.8175745, ...]
>
> Don't know, what this means, at the moment.
>
We have:
(V(2.0)-V(1.5))~*Z*F = [0.5000000, 1.692020, 5.401833, 16.42142, 48.08989, ... ]
+ (V(1.5)-V(1.0))~*Z*F = [0.5000000, 1.026262, 1.987223, 3.664115, 6.508259, ... ]
----------------------------------------------------------------------------------
= (V(2.0)-V(1.0))~*Z*F = [1.000000, 2.718282, 7.389056, 20.08554, 54.59815, ...]
As we should expect. But what are the values at "half-interval"-sums? Hmm.
Gottfried
Gottfried Helms
> Am 04.01.2010 13:33 schrieb Gottfried Helms:
>> What happens for fractional interval n-m? Examples:
>>
>> (V(3) -V(2.5))* Z * F = [0.5000000, 4.599387, 39.91445, 329.8331, ...]
>> (V(2) -V(1.5))* Z * F = [0.5000000, 1.692020, 5.401833, 16.42142, ...]
>> (V(1) -V(0.5))* Z * F = [0.5000000, 0.6224593, 0.7310586, 0.8175745, ...]
>>
>> Don't know, what this means, at the moment.
>>
>
> We have:
>
> (V(2.0)-V(1.5))*Z*F = [0.5000000, 1.692020, 5.401833, 16.42142, 48.08989, ... ]
> + (V(1.5)-V(1.0))*Z*F = [0.5000000, 1.026262, 1.987223, 3.664115, 6.508259, ... ]
> ----------------------------------------------------------------------------------
> = (V(2.0)-V(1.0))*Z*F = [1.000000, 2.718282, 7.389056, 20.08554, 54.59815, ...]
>
> As we should expect. But what are the values at "half-interval"-sums? Hmm.
>
(V(x) - V(x-1/2)) *Z*F =[ 0.5000000, f(x), ??? , ??? , ???, ... ]
are
f(x) = exp(x-1) *exp(1/2)/(1+exp(1/2))
so
f(2.0) = 1.69201988896
f(1.5) = 1.02626193950
Gottfried
Gottfried Helms
Ok, I assume it's a typo; I got identity using
Delta^-1 = I - 1/2 + 1/12 D - 1/720 D^3 + 1/30240 D^5 - 1/1209600 D^7
+ 1/47900160 D^9 - 691/1307674368000 D^11 + ...
where 1/1209600 D^7
instead of 1/129600 D^7
fits the scheme of zeta(-k)/k! *D^k
I used the likely correct value already in my other msgs.
Gottfried
mike3
<snip>
Interesting. V(x) is the Vandermonde vector, right?
I see this:
> (V(3) -V(2))* Z * F = [1.000000, 7.389056, 54.59815, 403.4288, 2980.958, ...]
> = [ exp(2)^0, exp(2), exp(2)^2, ...]
> = V(exp(2))
>
> (V(3) -V(1))* Z * F = [2.000000, 10.10734, 61.98721, 423.5143, 3035.556, .,...]
> = [ exp(1)^0 + exp(2)^0, exp(1)+ exp(2), exp(1)^2 + exp(2)^2, ...]
> = V(exp(1)) + V(exp(2))
...
> What happens for fractional interval n-m? Examples:
>
> (V(3) -V(2.5))* Z * F = [0.5000000, 4.599387, 39.91445, 329.8331, ...]
> (V(2) -V(1.5))* Z * F = [0.5000000, 1.692020, 5.401833, 16.42142, ...]
> (V(1) -V(0.5))* Z * F = [0.5000000, 0.6224593, 0.7310586, 0.8175745, ...]
>
> Don't know, what this means, at the moment.
If I got you right, (V(b) - V(a)) * Z * F gives a vector containing
sum_{n=a...b-1} exp(n)^m for integers m = 0, 1, 2, 3, ... So then
this should be fractional sum between fractional bounds. Yet we have
the formula sum_{n=0...x-1} exp(n) = (exp(x) - 1)/(e - 1) by
rearranging
the forward difference of exp(x), and if sum_{n=0...x-1} f(n) = F(x),
then F(b) - F(a) = sum_{n=a...b-1} f(n). Thus, using this, I get
F(3) - F(2.5) ~ 4.5993869175533570577789092497397393990
F(2) - F(1.5) ~ 1.6920198889607716531505372492840074049
F(1) - F(0.5) ~ 0.62245933120185456463890056574550847875
Which is really good, it agrees with your 2nd column, suggesting
that can be interpreted as the expected fractional summation of
exp(x) from 2.5 to 2, 1.5 to 1, 0.5 to 0. Now, does your *8th*
entry in the above converge? You only give the first 4 entries for
each (3.5, 2), etc. pair. This should sum exp(x)^7 = exp(7x). This
is a big one since it has exponential type greater than 2pi (somewhat
loosely, that means
it grows more strongly than exp(2pi x)). You should have the values
1167353.2698847691549395189455659224501 (3, 2.5)
1064.4883942390149291942016993796186346 (2, 1.5)
0.97068776924864368113471044188581841611 (1, 0.5)
(this is derived from F(x) = sum_{n=0...x-1} exp(7n) =
(exp(7x) - 1)/(e^7 - 1))
Then just calculate to at least the precision given in your
post, preferably more, I'd say to try for 16 digits precision.
Then try it using log(1+x), and tell me if the resulting 2nd
column agrees with the values for log(gamma(x+1)) if it
converges anywhere at all. You should get
F(1) - F(0.5) = 0.12078223763524522234551844578164721225
so for this you should expect to see something like (though
maybe with less decimals shown than here, depending on what
precision you run the calculation at)
(V(1) - V(0.5)) * Z * F = [ 0.5,
0.12078223763524522234551844578164721225, ... ]
. Does that happen? How about V(0.5) - V(0), if the above
has too small a convergence radius? It should have
-0.12078223763524522234551844578164721225 ...
Gottfried Helms
> it grows more strongly than exp(2pi x)). You should have the values
>
> 1167353.2698847691549395189455659224501 (3, 2.5)
> 1064.4883942390149291942016993796186346 (2, 1.5)
> 0.97068776924864368113471044188581841611 (1, 0.5)
>
> (this is derived from F(x) = sum_{n=0...x-1} exp(7n) =
> (exp(7x) - 1)/(e^7 - 1))
>
> Then just calculate to at least the precision given in your
> post, preferably more, I'd say to try for 16 digits precision.
>
However, their growth-rate seems to be geometric, the quotient
of c[k+2]/c[k] converges quickly to -1.24118449962...
so we can Euler-sum them. I show the coefficients [*1] and the
partial sums [*2] at the end of the msg
> Then try it using log(1+x), and tell me if the resulting 2nd
> column agrees with the values for log(gamma(x+1)) if it
> converges anywhere at all. You should get
>
> F(1) - F(0.5) = 0.12078223763524522234551844578164721225
Using PkPowSum(2.1,1.5) I get the partial sums up to index k=64
k partial sum
--------------------------------
...
61: 0.12078223 0895
62: 0.12078223 1514
63: 0.12078223 2077
64: 0.12078223 2582
--------------------------------
> 0.12078223 7635245222... <-- your value
but as I mentioned in a post some time ago, the PkPowSum is
somehow experimental extension of Eulersummation/N�rlund-means.
>
> so for this you should expect to see something like (though
> maybe with less decimals shown than here, depending on what
> precision you run the calculation at)
>
> (V(1) - V(0.5)) * Z * F = [ 0.5,
> 0.12078223763524522234551844578164721225, ... ]
>
> . Does that happen? How about V(0.5) - V(0), if the above
> has too small a convergence radius? It should have
> -0.12078223763524522234551844578164721225 ...
Using PkPowSum(2.1,1.5) I get the partial sums up to index k=64
...
61 -0.12078223 0895
62 -0.12078223 1514
63 -0.12078223 2077
64 -0.12078223 2582
-----------------------
> -0.12078223 763524522... <-- your value
---------------------------------------------
I'll see, whether I can use a better summation procedure. If it
is generally, that the occuring series approximate geometric
growth very quickly it should be easy to separate into an
exact part (the divergent geometric part which has then a closed form)
and a fast converging remainder.
Conclusion:
The conclusion of this so far is, that indeed we can model
your series using the matrices of my framework, and especially
the ZETA-matrix-variant resembles the operator you were thinking
about.
Perhaps there is some proceeding from here?
Gottfried
======================================================================
[*1] first few terms of series (dot-product) (V(1.0)-V(0.5)) * Z * F
0.500000000000
0.875000000000
0
-0.893229166667
0
1.09420572917
0
-1.35635918511
0
1.68325947891
0
-2.08920391116
0
2.59308316349
0
-3.21849403010
0
3.99474481976
0
-4.95821533884
0
6.15406002276
0
[*2] Partial (Euler-) sums for (V(1.0)-V(0.5)) * Z * F
partial sum up to k'th coefficient
k only correct digits displayed
----------------------------------------
...
56 0.9706877692486436811347**********
57 0.970687769248643681134710********
58 0.970687769248643681134710********
59 0.9706877692486436811347104*******
60 0.9706877692486436811347104*******
61 0.9706877692486436811347104******
62 0.9706877692486436811347104******
63 0.97068776924864368113471044*****
64 0.97068776924864368113471044*****
...
=======================================
64 0.97068776924864368113471044*****
> 0.97068776924864368113471044188581841611 (1, 0.5) <-- your value
=======================================